Question

Eight small drops of equal size are falling through air with a steady velocity of $10\text{cm/sec}$. If the drops coalesce, what would be the terminal velocity of the bigger drop?

• A. $40\text{cm/s}$
• B. $20\text{cm/s}$
• C. $30\text{cm/s}$
• D. $50\text{cm/s}$

Answer 1

The correct option is A $40\text{cm/s}$

Terminal velocity , $v_T=\frac{2r^2(\sigma -\rho )g}{9\eta }$
where $\sigma \to$ density of body falling through the given fluid of density $\rho$
$\Rightarrow v_T\propto r^2$
Let the terminal velocity for smaller drop ($r$) and bigger drop ($R$) be $v_T$ and $v_T^{\prime }$ respectively.
$\therefore \frac{v_T}{v_T^{\prime }}=\frac{r^2}{R^2}...\left(i\right)$

After the small drops coalesce, $\text{Volume of the big drop}=8\times \text{Volume of each small drop}$
$\frac{4}{3}\pi R^3=8\times \frac{4}{3}\pi r^3$
$\Rightarrow R=2r$
Given, terminal velocity (steady velocity) of small drop is $v_T=10\text{cm/s}$
Substituting in Eq.$\left(i\right)$,
$\frac{10}{v_T^{\prime }}=\frac{r^2}{(2r{)}^2}=\frac{1}{4}$
Terminal velocity of bigger drop will be
$v_T^{\prime }=4\times v_T=40\text{cm/s}$